I am interested in proving a formula for the integrated Chebyshev function $$\psi_1(x) = \int_1^x\psi(t)dt$$ that is a sum of all residues of the function $$f(s) = \frac{x^{s + 1}}{s(s + 1)}\frac{\zeta'(s)}{\zeta(s)}.$$ I know that $$\psi_1(x) = -\frac{1}{2\pi i}\int_{c - i\infty}^{c + i\infty}f(s)ds$$ for $c > 1$, and that you can let $c = 1$ to get a contribution of $\textrm{Res}_{s = 1}f(s) = x^2/2$ from the pole at $s = 1$. So to prove my formula I want to consider a rectangular contour whose right side is the line $\textrm{Re}(s) = 1$ and then show that the integral of $f(s)$ vanishes as you let the side lengths go to infinity.
What I wanted to do is use that $$\frac{\zeta'(s)}{\zeta(s)} = \frac{d}{ds}\log\zeta(s).$$ I can then get the necessary estimates for $f(s)$ by estimating $\zeta(s)$. But this would require a lower bound on $\zeta(s)$ to ensure that it's not going to $0$. Any suggestions?
EDIT: By taking the logarithmic derivative of the functional equation and estimating each term separately, I believe I have reduced the problem to estimating $\zeta'/\zeta$ on the parts of the contour lying in the critical strip, that is, on lines from $1 + iT$ to $iT$ and from $-iT$ to $1 - iT$.